A Tangled Tale, by Lewis Carroll

Answers to Knot 10

§ 1. The Chelsea Pensioners

Problem. — If 70 per cent have lost an eye, 75 per cent an ear, 80 per cent an arm, 85 per cent a leg: what
percentage, at least, must have lost all four?

Answer. — Ten.

Solution. — (I adopt that of Polar Star, as being better than my own.) Adding the wounds together, we get
70+75+80+85=310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10.

Nineteen answers have been received. One is “5” but, as no working is given with it, it must, in accordance with the
rule, remain “a deed without a name”. Janet makes it “35 7/10”. I am sorry she has misunderstood the question, and has
supposed that those who had lost an ear were 75 per cent of those who had lost an eye; and so on. Of course, on this
supposition, the percentages must all be multiplied together. This she has done correctly, but I can give her no
honours, as I do not think the question will fairly bear her interpretation. Three Score and Ten makes it “19 3/8” Her
solution has given me — I will not say “many anxious days and sleepless nights”, for I wish to be strictly truthful,
but — some trouble in making any sense at all of it. She makes the number of “pensioners wounded once” to be 310 (“per
cent,” I supposd): dividing by 4, she gets 77½ as “average percentage”: again dividing by 4, she gets 19 3/8 as
“percentage wounded four times”. Does she suppose wounds of different kinds to “absorb” each other, so to speak! Then,
no doubt, the data are equivalent to 77 pensioners with one wound each and a half-pensioner with a half-wound. And does
she then suppose these concentrated wounds to be transferable, so that 3/4 of these unfortunates can obtain perfect
health by handing over their wounds to the remaining 1/4? Granting these suppositions, her answer is right; or rather
if the question had been, “A road is covered with one inch of gravel, along 77½ per cent of it. How much of it could be
covered 4 inches deep with the same material?” her answer would have been right. But alas, that wasn’t the question!
Delta makes some most amazing assumptions: “let every one who has not lost an eye have lost an ear,” “let every one who
has not lost both eyes and ears have lost an arm.” Her ideas of a battlefield are grim indeed. Fancy a warrior who
would continue fighting after losing both eyes, both ears, and both arms! This is a case which she (or “it “?)
evidently considers possible.

Next come eight writers who have made the unwarrantable assumption that, because 70 per cent have lost an eye,
therefore 30 per cent have not lost one, so that they have both eyes. This is illogical. If you give me a bag
containing 100 sovereigns, and if in an hour I come to you (my face not beaming with gratitude nearly so much as when I
received the bag) to say, “I am sorry to tell you that 70 of these sovereigns are bad,” do I thereby guarantee the
other 30 to be good? Perhaps I have not tested them yet. The sides of this illogical octagon are as follows, in
alphabetical order: Algernon Bray, Dinah Mite, G. S. C., Jane E., J. D. W., Magpie (who makes the delightful remark,
“Therefore 90 per cent have two of something,” recalling to one’s memory that fortunate monarch with whom Xerxes was so
much pleased that “he gave him ten of everything”!), S. S G., and Tokio.

Bradshaw of the Future and T. R. do the question in a piecemeal fashion — on the principle that the 70 per cent and
the 75 per cent, though commenced at opposite ends of the 100, must overlap by at least 45 per cent; and so on. This is
quite correct working, but not, I think, quite the best way of doing it.

The other five competitors will, I hope, feel themselves sufficiently glorified by being placed in the first class,
without my composing a Triumphal Ode for each!

Class List.

I.

Old Cat. Polar Straw. Old Hen. Simple Susan. White Sugar.

II.

Bradshaw of the Future. T.R.

III.

§ 2. Change of Day

I must postpone, sine die, the geographical problem — partly because I have not yet received the statistics I am
hoping for, and partly because I am myself so entirely puzzled by it; and when an examiner is himself dimly hovering
between a second class and a third, how is he to decide the position of others?

§ 3. The Son’s Ages

Problem. — At first, two of the ages are together equal to the third. A few years afterwards, two of them
are together double of the third. When the number of years since the first occasion is two-thirds of the sum of the
ages On that occasion, one age is 21. What are the other two?

Answer. — 15 and 18.

Solution. — Let the ages at first be x, y, (x + y) Now, if a+b=2c, then (a-n) + (b-n)=2(c-n), whatever be
the value of n. Hence the second relationship, if ever true, was always true. Hence it was true at first. But it cannot
be true that x and y are together double of (x +y). Hence it must be true of (x +y), together with x or y; and it does
not matter which we take. We assume, then, (x +y) +x = 2y,. i.e. y = 2x. Hence the three ages were, at first, x, 2x,
3x, and the number of years since that time is two-thirds of 6x, i.e. is 4x. Hence the present ages are 5x, 6x, 7x. The
ages are clearly integers, since this is only “the year when one of my sons comes of age”. Hence 7x=21, x=3, and the
other ages are 15, 18.

Eighteen answers have been received. One of the writers merely asserts that the first occasion was 12 years ago,
that the ages were then 9, 6, and 3; and that on the second occasion they were 14, 11, and 8! As a Roman father, I
ought to withhold the name of the rash writer; but respect for age makes me break the rule: it is Three Score and Ten.
Jane E. also asserts that the ages at first were 9, 6, 3: then she calculates the present ages, leaving the second
occasion unnoticed. Old Hen is nearly as bad; she “tried various numbers till I found one that fitted all the
conditions”; but merely scratching up the earth, and pecking about, is not the way to solve a problem, O venerable
bird! And close after Old Hen prowls, with hungry eyes, Old Cat, who calmly assumes, to begin with, that the son who
comes of age is the eldest. Eat your bird, Puss, for you will get nothing from me!

There are yet two zeroes to dispose of. Minerva assumes that, on every occasion, a son comes of age; and that it is
only such a son who is “tipped with gold” Is it wise thus to interpret, “Now, my boys, calculate your ages, and you
shall have the money”? Bradshaw of the Future says “let” the ages at first be 9, 6, 3, then assumes that the second
occasion was 6 years afterwards, and on these baseless assumptions brings out the right answers. Guide future
travelers, an thou wilt; thou art no Bradshaw for this Age!

Of those who win honours, the merely “honourable” are two. Dinah Mite ascertains (rightly) the relationship between
the three ages at first, but then assumes one of them to be “6”, thus making the rest of her solution tentative. M. F.
C. does the algebra all right up to the conclusion that the present ages are 5z, 6z, and 7z; it then assumes, without
giving any reason, that 7z=21.

Of the more honourable, Delta attempts a novelty — to discover which son comes of age by elimination: it assumes,
successively, that it is the middle one, and that it is the youngest; and in each case it apparently brings out an
absurdity. Still, as the proof contains the following bit of algebra: “63=7x+4y;.’. 21 =x +4/7 of y,” I trust it will
admit that its proof is not quite conclusive. The rest of its work is good. Magpie betrays the deplorable tendency of
her tribe — to appropriate any stray conclusion she comes across, without having any strict logical right to it.
Assuming A, B, C, as the ages at first, and E as the number of the years that have elapsed since then, she finds
(rightly) the 3 equations, 2A=B, C=B + A, D = 2B. She then says, “Supposing that A=1, then B=2, C=3, and D=4. Therefore
for A, B, C, D, four numbers are wanted which shall be to each other as 1:2:3:4.” It is in the “therefore” that I
detect the unconscientiousness of this bird. The conclusion is true, but this is only because the equations are
“homogeneous” (i.e. having one “unknown” in each term), a fact which I strongly suspect had not been grasped — I beg
pardon, clawed — by her. Were I to lay this little pitfall: “A+1 =B, B+1 =C; supposing A = 1, then B =2, and C =3.
Therefore for A, B, C, three numbers are wanted which shall be to one another as 1:2:3,” would you not flutter down
into it, O Magpie! as amiably as a Dove? Simple Susan is anything but simple to me. After ascertaining that the 3 ages
at first are as 3:2:1, she says, “Then, as two-thirds of their sum, added to one of them, =21, the sum cannot exceed
30, and consequently the highest cannot exceed 15.” I suppose her (mental) argument is something like this: “Two-thirds
of sum, + one age, =21;.’. sum, +3 halves of one age, =31½. But 3 halves of one age cannot be less than 1 1/2 [Here I
perceive that Simple Susan would on no account present a guinea to a newborn baby!]; hence the sum cannot exceed 30.”
This is ingenious, but her proof, after that, is (as she candidly admits) “clumsy and round-about”. She finds that
there are 5 possible sets of ages, and eliminates four of them. Suppose that, instead of 5, there had been 5 million
possible sets! Would Simple Susan have courageously ordered in the necessary gallon of ink and ream of paper?

The solution sent in by C. R. is, like that of Simple Susan, partly tentative, and so does not rise higher than
being Clumsily Right.

Among those who have earned the highest honours, Algernon Bray solves the problem quite correctly, but adds that
there is nothing to exclude the supposition that all the ages were fractional. This would make the number of answers
infinite. Let me meekly protest that I never intended my readers to devote the rest of their lives to writing out
answers! E. M. Rix points out that, if fractional ages be admissible, any one of the three sons might be the one “come
of age but she rightly rejects this supposition on the ground that it would make the problem indeterminate. White Sugar
is the only one who has detected an oversight of mine: I had forgotten the possibility (which of course ought to be
allowed for) that the son who came of age that year, need not have done so by that day, so that he might be only 20.
This gives a second solution, viz., 20, 24, 28. Well said, pure Crystal! Verily, thy “fair discourse hath been as
sugar”!

Class List.

I.

II.

C. R. Magpie. Delta. Simple Susan.

III.

Dinah Mite. M. F. C.

I have received more than one remonstrance on my assertion, in the Chelsea Pensioners’ problem, that it was
illogical to assume, from the datum, “70 per cent have lost an eye,” that 30 per cent have not. Algernon Bry states, as
a paralel case, “Suppose Tommy’s father gives him 4 apples, and he eats one of them, how many has he left?” and says,
“I think we are justified in answering, 3.” I think so too. There is no “must” here, and data are evidently meant to
fix the answer exactly: but, if the question were set me, “How many must he have left?” I should understand the data to
be that his father gave him 4 at least, but may have given him more.

I take this opportunity of thanking those who have sent, along with their answers to the Tenth Knot, regrets that
there are no more Knots to come, or petitions that I should recall my resolution to bring them to an end. I am most
grateful for their kind words; but I think it wisest to end what, at best, was but a lame attempt. “The stretched metre
of an antique song” is beyond my compass; and my puppets were neither distinctly in my life (like those I now address),
nor yet (like Alice and the Mock Turtle) distinctly out of it. Yet let me at least fancy, as I lay down the pen, that I
carry with me into my silent life, dear reader, a farewell smile from your unseen face, “and a kindly farewell pressure
from your unfelt hand! And so, good night! Parting is such sweet sorrow, that I Shall say “good night!” till it be
morrow.

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